Easy problem about two bars connected by a spring

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In summary, your solution uses x1, x2 and x to calculate the distance between the bars, but if you exclude l0 then all your equations work. You can then add l0 back in to get the min and max values of the separation.
  • #1
LCSphysicist
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Homework Statement
All below.
Relevant Equations
All below.
1591824796655.png
1591824828055.png

My solution:
The distance between the block is x2-x1.
x2'' = F/m2 - k(x2-x1)/m2
x1'' = k(x2-x1)/m1

x2''-x1'' = x''.

x'' = F/m2 - kx(1/m1 + 1/m2)

Being y = (1/m1 + 1/m2)

That is>

x = (lo - Fy/m2k)*cos(wt) + Fy/m2k

xmin = lo
xmáx = -lo + 2*F*y/(m2*k) = -lo + 2*m1*F/k(m1+m2)

But the answer to xmax is lo + 2*m1*F/k(m1+m2)

Do you see any error?
 
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  • #2
The way you used x1, x2 and x, you effectively defined these as though l0=0. That's fine so long as you remember to add l0 back right at the end. But somehow you got it as being multiplied by the cos term.
 
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  • #3
haruspex said:
The way you used x1, x2 and x, you effectively defined these as though l0=0. That's fine so long as you remember to add l0 back right at the end. But somehow you got it as being multiplied by the cos term.

I am not sure if i understood, are you saying that maybe i made an error in resolving the differential equation?

x'' = F/m2 - kx(1/m1 + 1/m2)
x = Acos(wt) + Bsin(wt) + Fy/m2k `(actually 1/y = 1/m1 + 1/m2)
xo = lo = A + Fy/m2k
x'o = vo = 0 = Bw
B equals zero
A equals lo - Fy/m2k

So we get the final answer
(lo - Fy/m2k)*cos(wt) + Fy/m2k
?
 
  • #4
You use ##kx## for your force, but ##x## is the distance between the bars whilst ##x-l_0## is the extension! If you replace change your equations accordingly, you will get that the general solution is then $$x = A\cos`{(\omega t)} + \frac{F}{kym_2} + l_0$$ and you can deduce that ##A = -\frac{F}{kym_2}## with ##x(0) = l_0##, which then gives you the correct result.
 
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  • #5
Oh i betrayed myself>>>>>>>>>>>>>>>:O

thank you all.
 
  • #6
LCSphysicist said:
I am not sure if i understood, are you saying that maybe i made an error in resolving the differential equation?

x'' = F/m2 - kx(1/m1 + 1/m2)
x = Acos(wt) + Bsin(wt) + Fy/m2k `(actually 1/y = 1/m1 + 1/m2)
xo = lo = A + Fy/m2k
x'o = vo = 0 = Bw
B equals zero
A equals lo - Fy/m2k

So we get the final answer
(lo - Fy/m2k)*cos(wt) + Fy/m2k
?
I am saying that your solution was technically in error right at the start:
The distance between the block is x2-x1.
x2'' = F/m2 - k(x2-x1)/m2


If the distance between the blocks is x2-x1 then the tension is k(x2-x1-l0).
If you redefine your x variables so as to exclude l0 then all your equations work, but when you solve the ODE there will be no l0.
Having solved it, and obtained the min and max values of x, you can add l0 back to get the min and max values of the separation.
 
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  • #7
I wonder how many other ways there are of doing it. One might be to consider that ##a_{CM} = \frac{F}{(m_1 + m_2)}##. If we then transform into the frame of the CM, then both masses are performing SHM under the action of the inertial forces ##-\frac{Fm_1}{(m_1 + m_2)}## and ##-\frac{Fm_2}{(m_1 + m_2)}## respectively.

Once you work out the initial ratio of the lengths of spring on either side of the CM in terms of ##l_0##, then you can also work out the effective spring constants in terms of ##k##. And it's just like two springs in a gravitational field! You can quickly calculate the equilibrium positions and from that obtain the final result.

Maybe it takes a little longer than your method in this case; I'll try it tomorrow!
 
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Related to Easy problem about two bars connected by a spring

1. What is the purpose of studying the "Easy problem about two bars connected by a spring"?

The purpose of studying this problem is to understand the behavior of a simple mechanical system consisting of two bars connected by a spring. This problem is commonly used in introductory physics courses to introduce students to concepts such as equilibrium, Hooke's law, and simple harmonic motion.

2. How do the properties of the spring affect the behavior of the system?

The properties of the spring, such as its stiffness or spring constant, determine the amount of force required to stretch or compress the spring. This, in turn, affects the system's equilibrium position and the amplitude and frequency of its oscillations.

3. Can this problem be solved analytically or does it require numerical methods?

This problem can be solved analytically using principles of mechanics and calculus. However, for more complex systems or when considering real-world factors such as friction, numerical methods may be necessary.

4. What are the assumptions made in solving this problem?

The main assumptions made in solving this problem are that the bars are rigid and the spring is ideal, meaning it follows Hooke's law and has no mass or damping. Additionally, the problem assumes that the system is in a vacuum and there are no external forces acting on it.

5. How does this problem relate to real-world applications?

This problem is a simplified version of many real-world systems, such as a simple pendulum or a mass-spring-damper system. Understanding the behavior of these simple systems can help in understanding more complex systems and can be applied in various fields such as engineering, physics, and biology.

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